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I have the following equivalence relation problem.

$Let \ R\subseteq 2^S*2^S = \{(A,B):|A\cap T|=|B \cap T|\}\ where\ S=\{0,1,2\} \ and \ T=\{1,2\} \ Show \ that \ R \ is \ a \ equivalence \ relation \ to \ 2^S \ and \ describe \ the \ equivalence \ classes\ $

I understand that in order to show that $R$ is an equivalence relation I need to show that is $R$ is symmetric, reflexive and transitive.

Though I have two questions.

  1. What the result of $2^S$ will be?
  2. What steps do I need to follow to find the equivalence classes of an equivalence relation?
Wanderer
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  • I think you meant $;2^s= P(S)=$ the set of all subsets of $;S;$, and then $;R\subset 2^S\times 2^S;$ . What you mean by (1) is beyond my understanding, and about (2): there are only $;8;$ elements in $;2^S;$ so it shouldn't be that hard to do the actual partition of $;2^S;$ by equivalence classes. For example, $;[\emptyset]={\emptyset,,,{2}};$ – Timbuc Feb 08 '15 at 13:34
  • I thought 2^S to be power. I guess I missed the symbolism of the power Set. 1 is now understood. About 2 I still have a hard time understanding how you found [0]={0,{2}} – Wanderer Feb 08 '15 at 13:38
  • Nice, now try to separate all the elements in $;2^S=P(S);$ in different equivalence classes. – Timbuc Feb 08 '15 at 13:40
  • How can I do that? – Wanderer Feb 08 '15 at 13:46

1 Answers1

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I already gave you an example in the comment. For another one:

$$|\{1\}\cap T|=1\implies X\in\left[\,\{1\}\,\right]\iff |X\cap T|=1$$

so for example, we have in this case

$$\{1\}\,,\,\{0\}\,,\,\{0,2\}\,,\,\{1,2\}\in\left[\,\{1\}\,\right]$$

Now you try other cases.

Timbuc
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